# 26% of adults wear contact lenses and two are selected randomly. Given that at least 1 of these adults wears lenses...

$26\%$ of adults wear contact lenses and two are selected randomly. Given that at least $1$ of these adults wears lenses, what is the probability that both wear contact lenses?

So I figured $$P(A \text{ and } B) = P(A) P(B|A) = 0.26 \times 0.26 = P(0.26) P(B|A).$$ So I divide $0.0676$ by $0.26$ and get $0.26$. I'm not sure that I've done this right I've been looking for answers online and I've been stuck for an hour. Help is much needed. Thanks.

I also tried adding the sums of the possibilities. Probability of $A$ but not $B$. So $0.26 \times 0.76$ then (Prob of not $A$ but $B$) $+ 0.76 \times 0.26$ which gave me $\longrightarrow 0.1924 + 0.1924 = 0.3848$. Then divide $0.0676$ by $0.3848$? Sorry I don't know how to make proper math symbols.

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– Em.
Commented Jan 9, 2016 at 21:40
• It's so tempting to answer $0.26$. Commented Jan 9, 2016 at 21:45

26% of adults wear contact lenses and two are selected randomly. Given that at least 1 of these adults wears lenses

Let $A = \{\text{At least 1 wears lenses}\}$, $B = \{\text{2 wear lenses}\}$. Then \begin{align*} P(B|A) &= \frac{P(AB)}{P(A)}\\ &= \frac{P(B)}{P(\text{One of two wears})+P(\text{Two of two wear})} \\ &= \frac{.26(.26)}{\binom{2}{1}(.26)(1-.26)+\binom{2}{2}(.26)^2(1-.26)^0}\\ &= 0.1494253 \end{align*} since $A\cap B =B$.

You can also calculate as follows \begin{align*} P(B|A) &= \frac{P(AB)}{P(A)}\\ &= \frac{P(B)}{1-P(\bar A)} \\ &=\frac{.26(.26)}{1-\binom{2}{0}(.26)^0(1-.26)^2}\\ &=0.1494253. \end{align*}

• Why do you add the sum of the probability of Two of Two wear? Is it just because you have to take every circumstance then divide? Commented Jan 9, 2016 at 21:32
• $P(A)$ means the probability of at least one wears lenses. Then either one wears lenses or two wear lenses. So $P(A)$ is equal to the sum of the probability that exactly one wears or exactly two wears, since these events are mutually exclusive.
– Em.
Commented Jan 9, 2016 at 21:39
• Thank you very much for your help! I appreciate it very much. Commented Jan 9, 2016 at 21:44
• @TaylorKern No problem. Glad to help.
– Em.
Commented Jan 9, 2016 at 21:52